mos caps ii; mosfets i - mit opencourseware · pdf filelecture 10 - mos caps ii; ......

6.012 - Microelectronic Devices and Circuits

Lecture 10 - MOS Caps II;

p-Si

n+

B

SG

SiO2+–

vGS

(= vGB)

MOSFETs c

*

I - Outline

• Review - MOS Capacitor The "Delta-Depletion Approximation"

(n-MOS example)

Flat-band voltage: VFB ≡ vGB such that φ(0) = φp-Si: VFB = φp-Si – φm

Threshold voltage: VT ≡ vGB such that φ(0) = – φp-Si: VT = VFB – 2φp-Si + [2εSi qNA|2φp-Si| ]1/2/Cox

Inversion layer sheet charge density: qN* = – Cox

*[vGC – VT]

• Charge stores - qG(vGB) from below VFB to above VT Gate Charge: qG(vGB) from below VFB to above VT

Gate Capacitance: Cgb(VGB) Sub-threshold charge: qN(vGB) below VT

• 3-Terminal MOS Capacitors - Bias between B and C Impact is on VT(vBC): |2φp-Si| → (|2φp-Si| - vBC)

• MOS Field Effect Transistors - Basics of model Gradual Channel Model: electrostatics problem normal to channel

drift problem in the plane of the channel Clif Fonstad, 10/15/09 Lecture 10 - Slide 1

The n-MOS capacitor

Right: Basic device with vBC = 0

p-Si

n+

B

SG

SiO2+–

vGS

(= vGB)C

Below: One-dimensional structure for depletion approximation analysis*

Clif Fonstad, 10/15/09 Lecture 10 - Slide 2

BG

+ –

p-SiSiO 2

x-tox 0

vGB

* Note: We can't forget the n+ region is there; we will need electrons, and they will come from there.

MOS Capacitors: Where do the electrons in the inversion layer come from?

Diffusion from the p-type substrate? If we relied on diffusion of minority carrier electrons from

the p-type substrate it would take a long time to build up the inversion layer charge. The current density of elec-trons flowing to the interface is just the current across a reverse biased junction (the p-substrate to the inversion layer in this case):

!

Je = qni

2 De

NAwp,eff

[Coul/cm2

- s]

The time, τ, it takes this flux to build up an inversion charge

so τ is

!

"qN

* =#ox

tox

" vGB $VT( )is the the increase in the charge, Δqn, divided by Je:

!

" =#qN

*

Je

=$oxNAwp,eff

qni

2Detox

# vGB %VT( )


Diffusion from the p-type substrate. cont.? Using NA = 1018 cm-3, tox = 3 nm, wp,eff = 10 µm, De = 40 cm2/V

and Δ(vGB-VT) = 0.5 V in the preceding expression for τ we find τ ≈ 50 hr!

Flow from the adjacent n+-region? As the surface potential is increased, the potential energy

barrier between the adjacent n+ region and the region under the gate is reduced for electrons and they readily flow (diffuse in weak inversion, and drift and diffuse in strong inversion) into the channel; that's why the n+ region is put there:

There are many electrons here and they don't have far to go once the barrier is lowered.

Lecture 10 - Slide 4

p-Si

n+

B

SG

SiO2+–

vGS

(= vGB)

Clif Fonstad, 10/15/09

Electrostatic potential and net charge profiles - regions and boundaries

φXDT

p

φ(x)

-tox φ

φ(x) φ(x)

φp φ

vGB -φ

|2φp |-tox vGB -tox xd xxx m m φm

pφpvGB qNAXDT +

xd ox ox

ρ(x) ρ(x) qNAxd

ρ(x) Cox *(vGB - VT)

C

- C *(vGB - VFB) -t -tox XDT x-tox xx qD* = -qNAXDT −qNA*(vGB - VFB)ox

−qNA qD* = -qNAxdvGB qN

* = - Cox *(vGB - VT)

Acccumulation Depletion (Weak Inversion) Strong Inversionwhen φ(0) > 0 vGB < VFB VFB < vGB < VT VT < vGB

vGB

Flat Band Voltage Threshold Voltage|qNA)1/2/C– φ VT = VFB+|2φ |+(2εSi|2φ *

XDT

VFB = φp m

φ

p p ox φ(x)

φm φ

φ(x)

-φpvGB |2φ |p-tox -tox xx mvGB φp

p qNAXDT ρ(x) ρ(x)

-tox -tox XDT xx −qNA qD

* = -qNAXDT Clif Fonstad, 10/15/09 Lecture 10 - Slide 5

MOS Capacitors: the gate charge as vGB is varied


vGB [V] VT

VFB

qG* [coul/cm2]

qNAPXDT

!

qG

" = Cox

"vGB #VT( )

+ qNAP XDT

Inversion Layer

Charge

!

qG

"(vGB ) =

Cox

"vGB #VFB( ) for vGB $VFB

%SiqNA

Cox

"1+

2Cox

"2vGB #VFB( )%SiqNA

#1

&

' ( (

)

* + + for VFB $ vGB $VT

Cox

"vGB #VT( ) + qNA XDT for VT $ vGB

,

-

.

.

/

.

.

The charge expressions:

!

qG

" =#SiqNA

Cox

"1+

2Cox

"2vGB $VFB( )#SiqNA

$1

%

& ' '

(

) * *

DepletionRegionCharge

!

qG

" = Cox

"vGB #VFB( ) Accumulation

Layer Charge

!

Cox

"#$ox

tox

MOS Capacitors: the small signal linear gate capacitance, Cgb(VGB)

Cgb(VGB) [coul/V]

!

Cgb (VGB ) " A#qG

$

#vGB vGB =VBG

CAccumulation Inversion

VGB [V] VTVFB

ox Depletion

GThis expression can also be written

!

Cgb (VGB ) = Atox

"ox

+xd (VGB )

"Si

#

$ %

&

' (

)1εox Aε /t [= C ]ox ox oxas: εSi AεSi/xd(VGB)


!

Cgb (VGB ) =

A Cox

" for VGB #VFB

A Cox

"1+

2Cox

"2VGB $VFB( )%SiqNA

for VFB #VGB #VT

A Cox

" for VT #VGB

&

' ( (

)

( (

Lecture 10 - Slide 7B

MOS Capacitors: How good is all this modeling? How can we know?

Poisson's Equation in MOS As we argued when starting, Jh and Je are zero in steady

state so the carrier populations are in equilibrium with the potential barriers, φ(x), as they are in thermal equilibrium, and we have:

!

n(x) = nieq" (x ) kT

and p(x) = nie#q" (x ) kT

Once again this means we can find φ(x), and then n(x) and p(x), by solving Poisson's equation:

!

d2"(x)

dx2

= #q

$ni e

#q" (x ) / kT # eq" (x ) / kT( ) + Nd (x) # Na (x)[ ]

This version is only valid, however, when |φ(x)| ≤ -φp. When |φ(x)| > -φp we have accumulation and inversion layers,

and we assume them to be infinitely thin sheets of charge, i.e. we model them as delta functions.


Poisson's Equation calculation of gate charge Calculation compared with depletion approximation

model for tox = 3 nm and NA = 1018 cm-3:

tox,eff ≈ 3.2 nm

We'll look in this vicinity next.t

Clif Fonstad, 10/15/09 Lecture 10 - Slide 9 Plot courtesy of Prof. Antoniadis

ox,eff ≈ 3.3 nm

MOS Capacitors: Sub-threshold chargeAssessing how much we are neglecting

Sheet density of electrons below threshold in weak inversion: In the depletion approximation for the MOS we say that the

charge due to the electrons is negligible before we reach threshold and the strong inversion layer builds up:

!

qN ( inversion )vGB( ) = "Cox

*vGB "VT( )

But how good an approximation is this? To see, we calculate the electron charge below threshold (weak inversion):

!

qN (sub" threshold )vGB( ) = " q nie

q# (x ) / kTdx

xi vGB( )

0

$

This integral is difficult to do because φ(x) is non-linear

!

"(x) = "p +qNA

2#Si

x - xd( )2

but if we use a linear approximation for φ(x) near x = 0, where the term in the integral is largest, we can get a very good approximate analytical expression for the integral.


Sub-threshold electron charge, cont. We begin by saying

!

"(x) # "(0) + ax where a $d"(x)

dx x= 0

= %2qNA "(0) %"p[ ]

&Si

where

With this linear approximation to φ(x) we can do the integral and find

!

qN (sub" threshold )vGB( ) # q

kT

q

n(0)

a= " q

kT

q

$Si

2qNA %(0) "%p[ ]nie

q% (0) kT

To proceed it is easiest to evaluate this expression for various values of φ(0) below threshold (when its value is |φp|), and to also find the corresponding value of vGB, from

!

vGB "VFB = #(0) "#p +tox

$ox

2$SiqNA #(0) "#p[ ]This has been done and is plotted along with the strong inversion layer charge above threshold on the following foil.


Sub-threshold electron charge, cont.

6 mV

Neglecting this charge results in a 6 mV error in the threshold voltage value, a very minor impact. We will see its impact on sub-threshold MOSFET operation in Lecture 12.


MOS Capacitors: A few more questions you might haveabout our model

Why does the depletion stop growing above threshold? A positive voltage on the gate must be terminated on negativecharge in the semiconductor. Initially the only negative chargesare the ionized acceptors, but above threshold the electrons in the strong inversion layer are numerous enough to terminate all thegate voltage in excess of VT. The electrostatic potential at 0+ doesnot increase further and the depletion region stops expanding.

How wide are the accumulation and strong inversion layers? A parameter that puts a rough upper bound on this is the extrinsic Debye length

!

LeD " kT#Si q2N

When N is 1019 cm-3, LeD is 1.25 nm. The figure on Foil 11 seems to say this is ~ 5x too large and that the number is nearer 0.3 nm.*

Is n, p = nie±qV/kT valid in those layers? It holds in Si until |φ| ≈ 0.54 V, but when |φ| is larger than this Si becomes "degenerate" and the carrier concentration is so largethat the simple models we use are no longer sufficient and thedependence on φ is more complex. Thinking of degenerate Si as a metal is far easier, and works extremely well for our purposes.

Clif Fonstad, 10/15/09 Lecture 10 - Slide 13 * Note that when N = 1020 cm-3, LeD ≈ 0.4 nm.

Bias between n+ region and substrate, cont. Reverse bias applied to substrate, I.e. vBC < 0

vBC < 0

p-Si

n+

B

CG

SiO2+– vGC

vBC +

–

Soon we will see how this will let us electronically adjust MOSFET threshold voltages when it is convenient for us to do so.


φ(x) With voltage between substrate and channel, vBC < 0

Threshold: vGC = VT(vBC) with vBC < 0 vGB = VT(vBC)

|2φp| – vBC

-tox XDT(vBC < 0) x

φp

ρ(x) qNAXDT

XDT(vBC < 0) x-tox

−qNA

φm

-φp XDT(vCB = 0)

-φp – vBC

VT(vBC) = VFB + |2φp| + [2εSi(|2φp|-vBC)qNA]1/2/Cox *

{This is vGC at threshold}

XDT(vBC < 0) = [2εSi(|2φp|-vBC)/qNA]1/2

qN * = -qNAxDT

qN * = -[2εSi(|2φp|-vBC)qNA]1/2 Clif Fonstad, 10/15/09 Lecture 10 - Slide 15

Bias between n+ region and substrate, cont. - what electrons see

The barrier confining the electrons to the source is lowered by thevoltage on the gate, until high level injection occurs at threshold.


Bias between n+ region and substrate, cont. - what electrons see

The barrier confining the electrons to the source is lowered by thevoltage on the gate, until high level injection occurs at threshold.

When the source-substrate junction is reverse biased, the barrier ishigher, and the gate voltage needed to reach threshold is larger.


An n-channel MOSFET capacitor: reviewing the results of the

now allowing for vBC ≠ 0. depletion approximation,

vBC < 0

p-Si

n+

B

CG

SiO2+– vGC

vBC +

–

!

Flat - band voltage : VFB " vGB at which #(0) = #p$Si

VFB = #p$Si $ #m

Threshold voltage : VT " vGC at which #(0) = $#p$Si + vBC

VT (vBC ) = VFB $ 2#p$Si +1

Cox

*2%Si qNA 2#p$Si $ vBC[ ]{ }

1/ 2

Accumulation Depletion Inversion vCG

!

Inversion layer sheet charge density : qN

* = "Cox

*vGC "VT (vBC )[ ]

Accumulation layer sheet charge density : qP

* = "Cox

*vGB "VFB )[ ]

VFB VT0


An n-channel MOSFET

p-Si

B

G+vGS

n+

D

n+

S– vDS

vBS +

iG

iB

iD

Gradual Channel Approximation: There are two parts to the problem: the vertical electrostatics problem of relating the channel charge to the voltages, and the horizontal drift problem in the channel of relating the channel charge drift to the voltages. We will assume they can be worked independently and in sequence.


An n-channel MOSFET showing gradual channel axes

p-Si

B

G+vGS

n+

D

n+

S– vDS

vBS +

iG

iB

iD

L0y

x0

Extent into plane = W

Gradual Channel Approximation: - We first solve a one-dimensional electrostatics problem in the x direction

to find the channel charge, qN*(y)

- Then we solve a one-dimensional drift problem in the y direction to find the channel current, iD, as a function of vGS, vDS, and vBS.


We restrict voltages to the following ranges:

!

vBS " 0, vDS # 0

Gradual Channel Approximation i-v Modeling(n-channel MOS used as the example)

This means that the source-substrate and drain-substrate junctions are always reverse biased and thus that:

!

iB vGS ,vDS ,vBS( ) " 0

The gate oxide is insulating so we also have:

!

iG vGS ,vDS ,vBS( ) " 0

With the back current, iB, zero, and the gate current, iG, zero, the only current that is not trivial to model is iD.

The drain current, iD, is also zero except when when vGS > VT.

The in-plane problem: (for just a minute so we can see where we're going)

Looking at electron drift in the channel we write iD as

!

iD = "W sey (y)qn

*(y) = "W µe qn

*(vGS ,vBS ,vCS (y))

dvcs

dy (at moderate E - fields)

This can be integrated from y = 0 to y = L, and vCS = 0 to vCS = vDS, to get iD(vGS, vDS,vBS), but first we need qn

*(y).

Clif Fonstad, 10/15/09 Lecture 10 - Slide 21 (derivation continues on next foil)

Gradual Channel Approximation, cont. We get qn

*(y) from the normal problem, which we do next.

The normal problem: The channel charge at y is qn

*(y), which is:

!

qn

*(y) = "Cox

*vGC (y) "VT [vCS (y),vBS ]{ } = "Cox

*vGS " vCS (y) "VT [vCS (y),vBS ]{ }

with VT [vCS (y),vBS ] = VFB " 2#p"Si + 2$SiqNA 2#p"Si " vBS + vCS (y)[ ]{ }1/ 2

Cox

*

We can substitute this expression into the iD equation wejust had and integrate it, but the resulting expression is deemed too algebraically awkward:

!

iD (vDS ,vGS ,vBS ) =W

Lµe Cox

*vGS " 2#p "VFB "

vDS

2

$

% &

'

( ) vDS +

3

22*SiqNA 2#p + vDS " vBS( )

3 / 2

" 2#p " vBS( )3 / 2+

, - . / 0

1 2 3

4 5 6

A simpler, more common approach is to simply ignore the

!

VT [vCS (y),vBS ] "VT (vBS ) = VFB # 2$p#Si + 2%SiqNA 2$p#Si # vBS[ ]{ }1/ 2

Cox

*

dependence of VT on vCS, and thus to say

With this simplification we have:

!

qn

*(y) " #Cox

*vGS #VT (vBS ) # vCS (y)[ ]


Gradual Channel Approximation, cont.

The drain current expression (the in-plane problem): Putting our approximate expression for the channel charge

into the drain current expression we obtained from considering the in-plane problem, we find:

!

iD = "W µe qn

*(vCS )

dvCS

dy# W µe Cox

*vGS "VT (vBS ) " vCS (y){ }

dvCS

dy

This expression can be integrated with respect to dy for y = 0 to y = L. On the left-hand the integral with respect to y can be converted to one with respect to vCS, which ranges from 0 at y = 0, to vDS at y = L:

!

iD0

L

" dy = W µe Cox

*vGS #VT (vBS ) #vCS (y){ }dvCS

0

vDS

"

Doing the definite integrals on each side we obtain a relatively simple expression for iD:

!

iD L = W µe Cox

*vGS "VT (vBS ) "

vDS

2

# $ %

& ' (

vDS



The drain current expression, cont: Isolating the drain current, we have, finally:

!

iD (vGS ,vDS ,vBS ) =W

Lµe Cox

*vGS "VT (vBS ) "

vDS

2

#

$ % &

' ( vDS

Plotting this equation for increasing values of vGS we see that it traces inverted parabolas as shown below.

Note,however, that iD saturates at its peak value for larger values of vDS (solid lines); it doesn't fall off (dashed lines).

iD

vDS

inc.vGS



zero we find:

The drain current expression, cont: The point at which iD reaches its peak value and saturates is

easily found. Taking the derivative and setting it equal to

!

"iD

"vDS

= 0 when vDS = vGS #VT (vBS )[ ]

What happens physically at this voltage is that the channel (inversion) at the drain end of the channel disappears:

!

qn

*(L) " #Cox

*vGS #VT (vBS ) #vDS{ }

= 0 when vDS = vGS #VT (vBS )[ ]

For vDS > [vGS-VT(vBS)], all the additional drain-to-source voltage appears across the high resistance region at the drain end of the channel where the mobile charge density is very small, and iD remains constant independent of vDS:

!

iD (vGS ,vDS ,vBS ) =1

2

W

Lµe Cox

*vGS "VT (vBS )[ ] 2

for vDS > vGS "VT (vBS )[ ]


Gradual Channel Approximation, cont. The full model: Derived neglecting the

variation of the depletion layer charge with y.

!

iD (vGS ,vDS ,vBS ) =

0 for vGS "VT (vBS )[ ] < 0 < vDS

1

2

W

Lµe Cox

*vGS "VT (vBS )[ ]2

for 0 < vGS "VT (vBS )[ ] < vDS

W

Lµe Cox

*vGS "VT (vBS ) "

vDS

2

# $ %

& ' (

vDS for 0 < vDS < vGS "VT (vBS )[ ]

#

$

) ) )

%

) ) )

G B

S

D

+

–

+

vGSvBS

iD

iG iB

+

vDS

!

iG (vGS ,vDS ,vBS ) = 0

!

iB (vGS ,vDS ,vBS ) = 0

Cutoff

Saturation

Linear or Triode

Linear or Triode

iD

vDS

increasing

vGS

Cutoff Region Lecture 10 - Slide 26

Saturation or Region Forward Active

Region


The operating regions of MOSFETs and BJTs: Comparing an n-channel MOSFET and an npn BJT

MOSFET

G

S

D

+

––

+

vGS

vDS

iG

iD

iB

vBE vCE

iC

0.6 V 0.2 V

Forward Active RegionFAR

CutoffCutoff

Saturation

iC ! !F iBvCE > 0.2 ViB ! IBSe qVBE /kT

Input curve Output family

BJT

B

E

C

+

––

+

vBE

vCE

iB

iC

vDS

iD

Saturation (FAR)

Cutoff

Linearor

Triode

iD ! K [vGS - VT(vBS)]2/2!


6.012 - Microelectronic Devices and Circuits Lecture 10 - MOS Caps II; MOSFETs I - Summary

• Quantitative modeling (Apply depl. approx. to MOS cap., vBC = 0) Definitions: VFB ≡ vGB such that φ(0) = φp-Si Cox * ≡ εox/tox

VT ≡ vGB such that φ(0) = – φp-Si

Results and expressions (For n-MOS example) 1. Flat-band voltage, VFB = φp-Si – φm 2. Accumulation layer sheet charge density, qA* = – Cox *(vGB – VFB) 3. Maximum depletion region width, XDT = [2εSi|2φp-Si|/qNA]1/2

4. Threshold voltage, VT = VFB – 2φp-Si + [2εSi qNA|2φp-Si|]1/2/Cox * 5. Inversion layer sheet charge density, qN* = – Cox *(vGB – VT)

• Subthreshold charge Negligible?: Yes in general, but (1) useful in some cases, and (2) an

issue for modern ULSI logic and memory circuits

• MOS with bias applied to the adjacent n+-region Maximum depletion region width: XDT = [2εSi(|2φp-Si| – vBC)/qNA]1/2

Threshold voltage: VT = VFB – 2φp-Si + [2εSi qNA(|2φp-Si| – vBC)]1/2/Cox *

(vGC at threshold) • MOSFET i-v (the Gradual Channel Approximation) Vertical problem for channel charge; in-plane drift to get current


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